Question

Where on the rope do you expect the tension will be greatest? Where do you expect it will be the least? A uniform rope with length L and mass m is held at one end and whirled in a horizontal circle with angular velocity ω. You can ignore the force of gravity on the rope. (a) At a point on the rope a distance r from the end that is held, what is the tension F? (b) What is the speed of transverse waves at this point? (c) Find the time required for a transverse wave to travel from one end of the rope to the other.

Answer 1

Answer:

a)$F=\dfrac{\omega^2 m}{2L}\left({L^2}-{r^2}\right)$

b)$V=\omega \sqrt{\dfrac{L^2-r^2}{2}}$

c)$t=\dfrac{\pi}{\omega \sqrt2}$

Explanation:

Given that

Length =L

Mass = m

Force on elemental part

dF= dm ω² r

dm = m/L dr

dF= ω² r m/L dr

By integrating from  r to L

$F=\dfrac{m}{L} \int_{r}^{L}\omega^2 r \ dr$

$F=\dfrac{\omega^2 m}{L}\left(\dfrac{L^2}{2}-\dfrac{r^2}{2}\right)$

$F=\dfrac{\omega^2 m}{2L}\left({L^2}-{r^2}\right)$

Velocity V

$V=\sqrt{\dfrac{F}{\dfrac{m}{L}}}$

$V=\sqrt{\dfrac{\dfrac{\omega^2 m}{2L}\left({L^2}-{r^2}\right)}{\dfrac{m}{L}}}$

$V=\omega \sqrt{\dfrac{L^2-r^2}{2}}$

$\dfrac{dr}{dt}=V=\omega \sqrt{\dfrac{L^2-r^2}{2}}$

$\int_{0}^{L}\dfrac{dr}{\sqrt{L^2-r^2}}=\int_{0}^{t} \dfrac{\omega }{\sqrt2}dt$

$sin^{-1}1=\ \dfrac{\omega }{\sqrt2}t$

$\dfrac{\pi}{2}=\ \dfrac{\omega }{\sqrt2}t$

$t=\dfrac{\pi}{\omega \sqrt2}$